Comparison of two values

Question

I have to figure out the relation between the quantity $(0.9/1.1)^2 +(1.1/0.9)^2$ and 2. How can i do this without explicitly calculating the first value, by using some laws of exponents?

Answer 1

I think the easiest way to see it is to use the following inequality: For $x>0$, the following holds:

$$x + \frac{1}{x} \geq 2$$

with equality only if $x=1$. This you can see by multiplying out by $x$ and then you get the equivalent expression $x^2 +1 - 2x = (x-1)^2 \geq 0$, which is true (and again equality only holds if $x=1$).

In your case just plug in $x=\left(\frac{0.9}{1.1}\right)^2$, which is $>0$ and $\neq 1$, to get $x+\frac{1}{x} > 2$.

Answer 2

The expression has the form $\;x^2+\dfrac1{x^2}$. Now observe that, $$x^2+\dfrac1{x^2}\ge 2\iff x^2-2+\dfrac1{x^2}=\Bigl(x-\frac1x\Bigr)^2\ge 0.$$

Answer 3

hint. $\frac{0.9^2}{1.1^2}+\frac{1.1^2}{0.9^2}-2=\left(\frac{0.9}{1.1}-\frac{1.1}{0.9}\right)^2$